Question

7. The force between two electrons when placed in air equal to 0.5 times the weight of the electron.Find distance two electrons.​

Answer 1

AnswEr :

Distance of Seperation is 7.12 m

Explanation :

According to the Question,

The electrostatic force between the two electrons is 0.5 times the weight of one electron. In other words,the electrostatic repulsion of electron is 2 times the magnitude of weight of an electron

Here,

• Mass of an electron (M) = 9.1 × 10^-31 Kg

• Charge of an electron (Q) = 1.6 × 10^-19 C

• Acceleration due to gravity (g) = 10 m/s²

$\rule{300}{2}$

Now,

Let us first out the weight of the electron

$\tt W = Mg \\ \\ \leadsto \tt W = 9.1 \times 10^{-31} \times 10 \\ \\ \leadsto \tt W = 9.1 \times 10^{-30} \ N$

$\rule{300}{2}$

From Coulomb's Inverse Square Law : (For two identical charges)

$\tt F_E = \dfrac{KQ^2}{r^2}$

$\rule{300}{2}$

In accordance with the given relation,

$\tt F_E = 0.5 W \\ \\ \longrightarrow \tt \: \dfrac{2KQ^2}{r^2} = 9.1 \times {10}^{ - 30}$

(Putting the values)

$\longrightarrow \: \tt \dfrac{2 \times 9 \times 10 {}^{9} \times (1.6 \times {10}^{ - 19} ) {}^{2} }{ {r}^{2} } = 9.1 \times 10 {}^{ - 30} \\ \\ \longrightarrow \: \tt \: {r}^{2} = \dfrac{4608 \times \cancel{{10}^{9 - 40}} }{91 \times \cancel{ {10}^{ - 31} }} \\ \\ \longrightarrow \: \tt \: {r}^{2} = 50.637 \\ \\ \large{ \longrightarrow \: \boxed{ \boxed{ \tt \: r = 7.12 \: m}}}$

$\rule{300}{2}$

$\rule{300}{2}$